`int1/(sqrt(x)-root(3)(x))dx`

Apply integral substitution:`u=x^(1/6)`

`=>du=1/6x^(1/6-1)dx`

`du=1/6x^(-5/6)dx`

`du=1/(6x^(5/6))dx`

`6x^(5/6)du=dx`

`6(x^(1/6))^5du=dx`

`6u^5du=dx`

`int1/(sqrt(x)-root(3)(x))dx=int(6u^5)/(u^3-u^2)du`

`=int(6u^5)/(u^2(u-1))du`

Take the constant out,

`=6intu^3/(u-1)du`

Integrand is an inproper rational function as degree of numerator is more than the degree of the denominator,

So let's carry out the division,

`u^3/(u-1)=u^2+u+1+1/(u-1)`

`=6int(u^2+u+1+1/(u-1))du`

Apply the sum rule,

`=6(intu^2du+intudu+int1du+int1/(u-1)du)`

Apply the power rule and the common integer:`int1/xdx=ln|x|`

`=6(u^3/3+u^2/2+u+ln|u-1|)`

Substitute back `u=x^(1/6)`

and add a constant C to the solution,

`=6(1/3(x^(1/6))^3+1/2(x^(1/6))^2+x^(1/6)+ln|x^(1/6)-1|)+C`

`=2x^(1/2)+3x^(1/3)+6x^(1/6)+6ln|x^(1/6)-1|+C`

`=2sqrt(x)+3root(3)(x)+6root(6)(x)+6ln|root(6)(x)-1|+C`